What is the vertices 8-5x^2 ?

The vertices 8-5x^2 is Maximum (0,8)

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vertices 8-5x^2

vertices

8 - 5 x^{2}

Maximum (0, 8)

Solution steps

y = 8 - 5 x^{2}

Rewrite

y = 8 - 5 x^{2}

in the form

y = a x^{2} + b x + c

y = - 5 x^{2} + 8

The parabola parameters are:

a = - 5, b = 0, c = 8

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{0}{2 ( - 5 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 8

Therefore the parabola vertex is

(0, 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 5

Maximum (0, 8)

f (x) = 2 x^{2} - 4 x - 6

f (x) = 2 x^{2} - 4 x - 3

2 x^{2} - 1 + 5 x

y = - x^{2} + 18 x - 82

+ 2 (x + 1)^{2} - 3